Question
Answer
$$(a-5)(a^2-7a+7)=a^3-12a^2+42a-35$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(a-5)(a^2-7a+7)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}a^3-7a^2+7a-5a^2+35a-35 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}a^3-12a^2+42a-35\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{a-5}\right) $ by each term in $ \left( a^2-7a+7\right) $. $$ \left( \color{blue}{a-5}\right) \cdot \left( a^2-7a+7\right) = a^3-7a^2+7a-5a^2+35a-35 $$ |
| ② | Combine like terms: $$ a^3 \color{blue}{-7a^2} + \color{red}{7a} \color{blue}{-5a^2} + \color{red}{35a} -35 = a^3 \color{blue}{-12a^2} + \color{red}{42a} -35 $$ |
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